Exponents Questions

Exponents questions are provided here to practice various types of questions based on the concept of exponents and powers. These questions are designed according to the latest CBSE/ICSE syllabus for class 8. Practising these questions will help students prepare for their class tests and even final exams.

Let a be any arbitrary number, now a + a = 2a, a + a + a = 3a, and if we multiply it n times we write a + a + a + … n times = na. So this is about adding a number repeatedly, but if we multiply a number repeatedly like a × a = a², a × a × a = a³, and a × a × a × a × …n times = aⁿ, then it is called exponents. The number aⁿ is an exponent whose base is a and power is n, and we say it as “a raised to the power n”.

Learn more about exponents and powers.

Exponents Questions with Solution

Let us solve this practice worksheet on exponents, using the concept and laws of exponents.

Question 1: Simplify the following:

(i) (3/4)⁸ × (4/3) ⁵ (ii) (5/7)⁵ × (5/7) ^–6

Solution:

(i) (3/4)⁸ × (4/3) ⁵ = (¾)⁸ × (¾) ^–5

= (3/4)^{8 – 5} = (3/4) ³

= 27/48

(ii) (5/7)⁵ × (5/7) ^–6 = (5/7)^{5 – 6}

= (5/7) ^–1 = 7/5

Question 2: Express each of the following as rational numbers:

(i) (4/5)⁴ (ii) (64/81)^3/2 (iii) ( –2/5) ^–4

Solution:

(i) (4/5)⁴ = 4⁴/5⁴ = 256/625

(ii) (64/81)^3/2 = [{64/81}^1/2]³

= [8/9]³ = 512/729

(iii) ( –2/5) ^–4 = ( –5/2)⁴

= 625/16

Question 3: Simplify each of the following:

(i) [(–2/3) ^–3 × ( –3/5)⁶] ÷ (2/3)⁵

(ii) (3/4) ⁵ ÷ (81)² × (5/3)³

(iii) [{( –2/3)² } ^–2] ^–1

Solution:

(i) [(–2/3) ^–3 × ( –3/5)⁶] ÷ (2/3)⁵

= [(–3/2) ³ × ( –3/5)⁶] × (3/2)⁵

= [3^{3 + 6 + 5}/(2^{3 + 5} × 5⁶)]

= 2^–8 3¹⁴ 5 ^–6

(ii) (3/4) ⁵ ÷ (81)² × (5/3)³

= (3/4) ⁵ ÷ (3⁴)² × (5/3)³

= (3/4) ⁵ × 3 ^–8 × (5/3)³

= (3 ^{5 –8 –3} × 5³)/4⁵

= 3 ^–6 4 ^–5 5 ³

(iii) [{( –2/3)² } ^–2] ^–1

= ( –2/3) ^{2 × ( –2) × –1}

= ( –2/3) ⁴

= 16/81.

Question 4: A number when divides ( –15) ^–1 results ( –5) ^–1. Find the number.

Solution:

Let x be the number such that

( –15) ^–1 ÷ x = ( –5) ^–1

⇒ –1/15 ÷ x = –⅕

⇒ –1/15 × 1/x = –⅕

⇒ –1/15x = –⅕

⇒ 15x = 5

⇒ x = ⅓ or 3 ^–1

Question 5: Simplify:

Solution:

We have

Question 6: Find the value of p, if 25^{(p – 1)} + 100 = 5^{(2p – 1)}.

Solution:

We have 25^{(p – 1)} + 100 = 5^{(2p – 1)}

⇒ 5^{(2p – 1)} – (5²)^{(p – 1)} = 100

⇒ 5^{(2p – 1)} – 5^{(2p – 2)} = 100

⇒ 5^{(2p – 2)} × (5 – 1) = 100

⇒ 5^{(2p – 2)} × 4 = 100

⇒ 5^{(2p – 2)} = 25

⇒ 5^{(2p – 2)} = 5²

On comparing the powers,

2p – 2 = 2 ⇒ p = 4

Question 7: Solve for x: 4^x + 2^x – 20 = 0

Solution:

We have 4^x + 2^x – 20 = 0

⇒ (2²)^x + 2^x – 20 = 0

⇒ 2^x + 2^x – 20 = 0

Put 2^x = t, the above equation becomes a quadratic equation

t² + t – 20 = 0

⇒ (t + 5)(t – 4) = 0

∴ We get, t = –5 or 4, that is;

2^x = – 5 or 2^x = 4

Since 2^x cannot be negative for any real number x,

2^x = 4 = 2²

⇒ x = 2 (on comparing the powers on both sides)

Question 8: Solve for x: 4^x – 6(2^x) + 8 = 0

Solution:

We have 4^x – 6(2^x) + 8 = 0

⇒ 2^2x – 6(2^x) + 8 = 0

Put 2^x = t, we get

t² – 6t + 8 = 0

⇒ (t – 4)(t – 2) = 0

∴ We get, t = 2 or 4, that is;

2^x = 2 or 2^x = 4

⇒ x = 1 or 2 (on comparing the powers on both sides)

Question 9: Solve of x: (3/5)³ × (3/5) ^–6 = (3/5) ^{2x – 1}

Solution:

We have, (3/5)³ × (3/5) ^–6 = (3/5) ^{2x – 1}

⇒ (3/5)^{3 – 6} = (3/5) ^{2x – 1}

Comparing powers on both sides,

⇒ 2x – 1 = –3

⇒ x = –1

Question 10: Simplify: (2^{m + 2} – 2^m)/2^m

Solution:

(2^{m + 2} – 2^m)/2^m = 2^m (2² – 1)/2^m

= 4 – 1 = 3.

Video Lesson on Exponents

Practice Questions on Exponents

1. Find: {(3/2) ^–1 + ( –2/5) ^–1}

2. Simplify:

3. Solve for x:

(i) 7^{x + 1} = 343

(ii) 3 ^{–(x + 1)} = 243

(iii) 49^x + 1 = 2(7^x)

4. Simplify: (14² – 13²)^5/3

5. Find the value of p: 3^{p + 8} = 27 ^{2p + 1}.

Keep visiting BYJU’S to get more such Maths lessons in a simple, concise and easy to understand way. Also, register at BYJU’S – The Learning App to get complete assistance for Maths preparation with video lessons, notes, tips and other study materials.